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I know that given say 4 quantum numbers $J^2$, $J_z$, $J_1^2$, $J_2^2$ (e.g. for the Hamiltonian $H=\lambda J_{1}.J{_2}$), the state |$J$,$J_z$,$J_1^2$,$J_2^2$>=|2,2,1,1> will be an energy eigenstate (by definition).

However, can there still be quantum states that have fixed energy (so are energy eigenstates) but not fixed values of all the 4 quantum numbers?

In other words, are you allowed to label an energy eigenstate with quantum numbers like |2, Jz, 1, 1> where $J_{z}$ is a superposition of, say, 1 and 2?

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  • $\begingroup$ "Allowed"? Who is going to stop you? What do you propose to do with that state? $\endgroup$ – Cosmas Zachos Apr 13 at 13:49
  • $\begingroup$ Thanks, I was just assuming before that every energy eigenstate had to be in the basis rather than a superposition of basis vectors. But I was wrong. $\endgroup$ – Alex Gower Apr 13 at 14:08
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What I found in the end is that it is not required for every energy eigenstate to be in the basis constructed using good quantum numbers, so long as you can still define the energy eigenstates using combations of these basis vectors.

E.g. the basis constructed of |px, py, pz> for free particles does not include the fixed energy eigenstate composed of a superposition of two equal magnitude opposite direction momentum eigenstates.

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